Backgammon Theory

Theory

Undefined equity

From:   Paul Tanenbaum
Address:   ptanenbaum@hotmail.com
Date:   15 August 1997
Subject:   Extremely theoretical
Forum:   rec.games.backgammon
Google:   5t2nr3$7bc$1@orthanc.reference.com

The diagram below should pique the interest of backgammon theoreticians. Its invention was motivated by the paper: "Optimal Doubling in Backgammon", E. Keeler & J. Spencer, Operations Research, Vol. 23, No.6. It is presented as a problem: What makes this position extraordinary? Those without access to the aforementioned paper will find this a real challenge, though there are some contributors to this newsgroup who are evidently perspicacious enough to suss it out on their own. I have since been informed that Bob Floyd published a position essentially identical in an obscure, now defunct, backgammon magazine several years ago. He therefore deserves recognition of first discovery. This posting is submitted for the edification of those unfamiliar with that article. Do any other positions exist which possess the same property? It should be noted that the Jacoby rule regarding an initial double may or may not be in effect; the conclusion is unaltered. Also, realize that it is quite legal. Money game X on roll - or O on roll - it doesn't matter! O X X ^ ^ ^ | | X O O O O O | | O O O O O | | | X | --- | | | 1 | | O | --- | | | | X X X X X X ^ O O ^ ^ | | O X X X X X _________________________________________________ < X home board > Paul Tanenbaum -- Posted using Reference.COM http://www.reference.com Browse, Search and Post Usenet and Mailing list Archive and Catalog. InReference, Inc. accepts no responsibility for the content of this posting.

Gary Wong writes:

Hmmm... looks like the first player to roll a 6 gets a tremendous advantage (either an anchor and 2 of the opponent on the bar against a 5 point board for 6-1 to 6-5, or 3 of the opponent on the bar for 6-6). And I guess 11/36 probability of this happening makes the situation a double/take for whoever's on roll, right? So the cube gets turned every roll until somebody rolls a 6? In that case, the expected gain of the player on roll (from this roll only) is 11/36 of 2 points (after doubling, ignoring gammons and backgammons) for 0.61 points; the expected gain of the opponent on the turn after is 25/36 (the first player missing) x 11/36 (the second player hitting) x 4 (the cube is turned twice) for 0.85 points; thereafter the expected gains continue increasing at a factor of 2 (for the cube) x 25/36 (for the probability of the game lasting that long) = 50/36 = 1.39. So the total expected gain is the difference of the odd or the even terms (depending which player you mean) of the geometric series a = 22/36, r = 50/36. But since r > 1, the series never converges, and so the expected gain for both players is infinite, right? (and so is the expected loss!) That sure counts as "extraordinary" in my book :-) Cheers, Gary. (GaryW on FIBS) -- Gary Wong, Computer Science Department, University of Auckland, New Zealand gary@cs.auckland.ac.nz http://www.cs.auckland.ac.nz/~gary/

Brian Sheppard writes:

Expected gain is infinite? Not at all! The expected gain in a backgammon game is never more than triple the cube. I don't know where you made your error, but you can rest assured that there are some. The way to analyze this type of position is with a "recurrent equation." In this type of equation the quantity you wish to evaluate comes up as a term on both sides of the equation after you analyze a few rolls. In this case, owing to the symmetry of the position, we obtain a "recurrence" of the quantity after just one roll. Let's analyze the cubeless case first. Call the equity of the side to move X, and let the equity of the side to move given that he rolls a 6 be E. Then we have the following equation: In 36 rolls, the side to move will have 11 sixes, winning E each time, and 25 misses, in which case he loses X, since the situation is exactly reversed. Therefore, 36*X = 11*E - 25*X So X = E * 11/61. In words: multiply the value of the position after a 6 by 11/61 to obtain the equity for the side to move. When we take the cube into account that changes things slightly. Let Y be the equity after the opponent takes. In 36 rolls the side to move will win E in 11 rolls, and will lose 2Y in 25 rolls (because the opponent will redouble). Now the equation is 36*Y = 11*E - 25 * 2Y, so Y = E * 11/86. To sum up: the side to roll should double, to raise his equity from 11/61 * E to 2 * 11/86 * E. The other side should take because 11/61 * E is definitely less than 1. Brian

Bob Koca writes:

Gary's only error, and it is a small one, was calling the expected value infinite. Undefined is more appropriate. Brian's recursion method normally works, however it depends on the expected values actually existing. Here they don't. Many people find examples of this sort confusing and I believe some of that stems from not knowing exactly what expected value means. Not getting overly technical, expected value can be thought of as the long term average. For example let's look at a very simple case. You have two checkers on the 3 point, opponent has 2 on ace point and the cube is centered. The expected gain to the player on roll is - 1/18. What this means is that as more and more games are played from this starting position the average gain to the player on roll will eventually get closer and closer to -1/18. (The exact mathematical statement is basically this but rigorously deals with "eventually" and "gets closer and closer") Suppose we played Gary's position many many times. The average winnings will NEVER settle down to a limit. The reason is that there will be eventually huger and huger amounts of points lost on a single game. Enough to knock off whatever progress was made towards the average gain approaching a limit. As an aside, i've seen several posts in the past stating that equities exist for backgammon and describing a recursive method to calculate them. This only works IF positions such as Gary's will never occur. The concept of undefined expected values is a little tricky and a little counterintuitive. If you are interested in exploring it further I would suggest looking at definitions of expected value, the weak and strong laws of large numbers, and the cauchy distribution (an example in which expected value does not exist, even though the distribution is symmetric). All of these will probably be in a good undergraduate level probability text. Bob Koca bobk on FIBS

Brian Sheppard writes:

First I want to tender my apologies to Gary, who it seems has grokked the real issue. I can get my head around the concept of "undefined expected values" for only moments at a time, and then it slips away. Perhaps if you could answer a few questions it might help me to understand what is going on. First: am I correct to say that the situation arises because the payoff is increasing at a factor of 2 per turn (because of the cube-turn) but the probability of that payoff is decreasing at the rate 25/36 (which is greater than 1/2)? The point is that the chance of a huge payoff is decreasing, but the payoffs are increasing faster. Second: suppose we take a sample of outcomes from this game, and observe the average amount won as time goes on for an indefinite number of games. Given an arbitrary positive number N, is it true that the series of averages will eventually be larger than N? Is it also true that the series of averages will eventually be less than -N? Third: is it correct to double? If it is not correct to double, then the position does have a theoretical equity. (But, what is the basis for deciding when to double if there is no equity to reason about?) Fourth: is it correct to take? If it is not correct to take, then the position does have theoretical equity. Fifth: If doubling and taking is correct, then this position has undefined equity. That means that positions that lead to this have undefined equity and so on. What process, if any, prevents backgammon as a whole from having undefined equity? Sixth: I am curious about computer evaluation of this situation. Does JF double for the side to roll? Thanks in advance to anyone who can answer any of these. Brian

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